The universal gravitational constant G is a physical constant found in Newton's Law of Gravitation. The approximate value of G is m3 G = 6.674 × 10-11 ~ (1) [kg · s² What is the conversion of G to the natural units?
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![The universal gravitational constant G is a physical constant
found in Newton's Law of Gravitation. The approximate value of G is
m3
G = 6.674 x 10-11
kg · s2
~ (1)
What is the conversion of G to the natural units?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc33efcf-8cc5-4c6e-b7a0-2fbe416d0a49%2F654a4526-130b-4822-8fcf-4117813e41c2%2Fdsxp08f_processed.png&w=3840&q=75)
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- I need help with a math question: Given that x = position, v = velocity, a = acceleration, m = mass. Which of the following will have units of kg * m/s^2 and why? m ∫ a dt m ∫ v dt 1/2m dv^2/dx m ∫ x dt 1/2m dv^2/dt m dx/dt m dv/dt m da/dt 1/2m ∫ v^2 dt I am including an image/reference table. Thanks!You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem: mearth = 5.9742 x 1024 kgrearth = 6.3781 x 106 mmmoon = 7.36 x 1022 kgrmoon = 1.7374 x 106 mdearth to moon = 3.844 x 108 m (center to center)G = 6.67428 x 10-11 N-m2/kg2 1) On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get? 2) Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is…Asap plzzxx